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skelet666 [1.2K]
3 years ago
9

Lance armstrong biking over 40 miles in one day would be considered muscular strength. true or false

Physics
1 answer:
Sonja [21]3 years ago
4 0

True.

Reason:

Muscular strength is the maximum force exerted by a muscle against some form of resistance.

In this example, lance Armstrong is continuously working against the gravitational force. Thus, it can be considered as muscular strength.

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A hockey puck is traveling to the left with a velocity of v=10 when it is struck by the hockey stick
Lera25 [3.4K]
We have to calculate the impulse of a hockey puck.
Imp = m * ( v 1 - v 2 ) = m * Δ v
v 1 = - 10 i m/s,
v 2 = ( 20 * cos 40° ) i + ( 20 * sin 40° ) j =
= ( 20 * 0.766 ) i + ( 20 * 0.64278 ) j = ( 15.32 i + 12.855 j ) m/s
Δ v = ( 15.32 i + 12.855 j ) - ( - 10 i ) =
= 15.32 i + 12.855 j + 10 i = 25.32 i + 12.855 j
| Δv | = √ ( 25.32² + 12.855²) = √806.35 = 28.4 m/s
Imp = 0.2 kg * 28.4 m/s = 5.68 N-s
Answer: D ) 5.68 N-s. 
 
4 0
3 years ago
Please help!!~
lisov135 [29]

Answer:

C. Waves transfer energy, but not matter.

Explanation: hope this helps :)

7 0
3 years ago
A line _____ on a typical Speed vs. Time graph means an object is experiencing a constant acceleration.
Savatey [412]

Answer:

The answer should be C. slanted upward to the right.

Hope this helps. :-)

7 0
3 years ago
Read 2 more answers
Why do we consider eyes, ears, and fingers as physical sensor​
seraphim [82]

Answer:

because

Explanation:

because we need eye to see if you don't have eyes how you regnise some one

if you have ears how can you know who's voice is who's

if don't have fingers how can you know what your are holding this is why they are called physical sensors

4 0
3 years ago
Read 2 more answers
A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o
Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

\delta = \frac{PL}{AE}

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

P = 68*10^3 N

E = 200GPa  

A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2

Replacing we have,

\delta = \frac{PL}{AE}

\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}

\delta = 0.001932m

\delta = 1.93mm

Therefore the change in length is 1.93mm

7 0
3 years ago
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